Algebraic manipulation: Factorise 9x^2 -1; hence, factorize 899, 8.99 x 10^4

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bumblebee123

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Hello,

I wasn't sure which topic to post this one under so I went with Beginner Algebra I hope that's ok.

I wonder if someone can help. I have a question as follows:

a)Factorise 9x^2 -1

Which I factorize correctly as:
(3x+1)(3x-1)

The next question is

b) Hence express as a product of its prime factors

i)899
ii)8.99 x 10^4


I find answer i) 899 by literally dividing until I get to the answer 29x31

the answer for ii) from the book is:

2x2x5x5x29x31
Though I don't know how to get the answer myself.

My question is how do you get the answers topart b i and ii?
The fact it says 'hence' makes me feel like I should use the answer to part 'a'. However, despite playing with it, I don't know how to use it or even why it would relate.
I can't find the answer to part bii myself. How would you find the answer to part b ii?

Thank you in advance for your help
 
Don't know why 'hence' is used. Bad translation?

Regarding b ii - you know what to do with 899. Can you rewrite 8.99 * 10^4 as 899 * X? Then you'll just need to factor X.
 
bumblebee123 said:
Hello,

I wasn't sure which topic to post this one under so I went with Beginner Algebra I hope that's ok.

I wonder if someone can help. I have a question as follows:

a)Factorise 9x^2 -1

Which I factorize correctly as:
(3x+1)(3x-1)

The next question is

b) Hence express as a product of its prime factors

i)899
ii)8.99 x 10^4


I find answer i) 899 by literally dividing until I get to the answer 29x31

the answer for ii) from the book is:

2x2x5x5x29x31
Though I don't know how to get the answer myself.

My question is how do you get the answers topart b i and ii?
The fact it says 'hence' makes me feel like I should use the answer to part 'a'. However, despite playing with it, I don't know how to use it or even why it would relate.
I can't find the answer to part bii myself. How would you find the answer to part b ii?

Thank you in advance for your help
Click to expand...
I am not sure that I see a great deal of point to the exercise.

However, with respect to bi

\(\displaystyle 899 = 900 - 1 = 9 * 10^2 - 1 = (3 * 10 - 1)(3 * 10 + 1) = 29 * 31.\)

And it is simple to determine that 29 and 31 are prime. It is a short-cut with respect to the sieve of Eratosthenes for any number that is recognized as being a difference of perfect squares.

For bii, you can use the result for bi.

\(\displaystyle 8.99 * 10^4 = 899 * 10^2 = 29 * 31 * (2 * 5)^2 = 2^2 * 5^2 * 29 * 31.\)

In short,

\(\displaystyle a,\ b,\ c \in \mathbb Z^+ \text { and } a = b^2 - c^2 \implies a = (b - c)(b + c).\)

That means that you can find the prime factors of a by finding the prime factors of b - c and b + c, which should be quicker than applying the sieve to a. But of course to use this trick, you first have to find b and c if they exist.
 
Last edited:
lev888 said:
Ok, now I see what 'hence' is about :)
Click to expand...
Yes. If the word "hence is used in part (b) of a question, it means you MUST use your result from part (a) to complete the question. If you don't, you would probably be marked wrong. Usually "hence, or otherwise" is used if an alternative approach is acceptable.
 
bumblebee123 said:
i)899
ii)8.99 x 10^4

I find answer i) 899 by literally dividing until I get to the answer 29x31

the answer for ii) from the book is:

2x2x5x5x29x31
Though I don't know how to get the answer myself.
Click to expand...

899=29x31
8.99 x 10^4=899x100
=29x31x10x10
=29x31x2x5x2x5
 
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